Calculus II Class 04/25

The force of air resistance on a falling body depends on its velocity v.  The acceleration of the body is related to the net force acting on it by F = m a = m dv/dt. 

So we see that dv/dt will depend in some way on v itself, which will give us an interesting differential equation.

In some circumstances (though not usually) the force of air resistance can be proportional to the velocity of the object.

In this case the net force is equal to the sum of the force exerted by gravity and the force of air resistance.  If we take the downward direction as positive then these forces are m g and - k v.

It follows that

m dv/dt = m g - k v.

This is a separable differential equation.  If we divide both sides by m we get

dv/dt = g - k / m * v.

We rearrange this first to the form

dv = (g - k / m * v) dt.

Dividing both sides by g - k / m * v we get

dv / (g - (k/m) v) = dt.

We integrate the left-hand side by substituting u = g - (k/m)v, obtaining du = -k/m * dv so that dv = -m/k * du.  Our integrand is therefore -m / k * (1/u) with antiderivative -m/k ln| u | = -m/k ln | g - (k/m) v |.

Our equation becomes

-m/k ln | g - (k/m) v | = t + c or

ln | g - (k/m) v | = -k/m(t + c) so that

g - (k/m) v = A e^(-k/m * t).  Solving for v we get

v = A e^(-k/m * t) + m g / k.

Substituting initial conditions and the value of k we can find A.  For initial velocity 0 the solution is

v = m g / k ( 1 - e^(-k/m * t) ).